2020 AMC 8 Problem 20

Attempt Problem 20 of the 2020 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

A scientist walking through a forest recorded as integers the heights of 55 trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
T100‾ mT211 mT300‾ mT400‾ mT500‾ mT‾00‾.2 m\begin{array}{|c|c|}\hline T_1&\underline{\phantom{00}}\,\mathrm m\\T_2&11\,\mathrm m\\T_3&\underline{\phantom{00}}\,\mathrm m\\T_4&\underline{\phantom{00}}\,\mathrm m\\T_5&\underline{\phantom{00}}\,\mathrm m\\\hline\overline T&\underline{\phantom{00}}.2\,\mathrm m\\\hline\end{array}
22.222.2
24.224.2
33.233.2
35.235.2
37.237.2
Answer: B
Concepts:caseworkmean
Small hint:
Integer heights force the neighbors of 1111 to be 2222
Big hint:
The average ending in 0.20.2 means the total ends in 11 mod 55
Video solution:
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Written solution:
Tree 22 is 1111 meters tall. Since all heights are integers and neighboring trees differ by a factor of 2,2, trees 11 and 33 must both be 2222 meters tall.
The first three trees total 22+11+22=5522+11+22=55 meters. The possible integer pairs for trees 44 and 55 are (11,22),(11,22), (44,22),(44,22), and (44,88).(44,88).
These give averages 17.6,17.6, 24.2,24.2, and 37.4,37.4, respectively. The notebook shows that the average ends in 0.2,0.2, so it must be 24.2.24.2.
Thus, the correct answer is B.
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